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Problem Set 5.3 | Universal Gravitation |
Jonathan Corbett |
25 March 2025 |
Problem Set | Universal Gravitation
1. A 70.0 kg human standing on Earth
Calculating the force of attraction between the human and Earth's center of gravity requires G, (a) Earth's radius, and the masses of the two bodies. As the human is standing on the surface of the Earth, the distance between their centers of mass is simply Earth's radius (ignoring the height of the human)
- a. Calculating Fg between the two masses:
|3 sf|$F_g = G \frac{m \dot M}{r^2}$|
|-:|:-|
|$m = 70.0 kg$
$M = 5.970x10^{24} kg$
$r = 6\,378x10^3 m$|$F_g = 6.673x10^{11} \left($ \frac{(70.0 kg)(5.970x10^{24}kg)}{(6\,378x10^3 m)^2} \right) $
$F_g = 685.66 N$
$F_g = 686 N$|
- b. Calculate the acceleration of the human
The acceleration of the human is centripetal, toward Earth's center.
| 3 sf |
Fg=m⋅ag |
m=70.0kg Fg=685.66N r=6378x103m |
ag=mFg ag=70.0kg685.66N ag=9.795142857...s2m ag=9.80s2m |